The sum of the series: $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots \ldots$. upto 10 terms, is:

Question

Aicharya · Accepted Answer

$\mathrm{T}_r=\frac{1}{1+2+3+\ldots+r}=\frac{2}{r(r+1)}$ $\mathrm{S}_{10}=2 \sum_{r=1}^{10} \frac{1}{r(r+1)}=2 \sum_{r=1}^{10}\left[\frac{r+1}{r(r+1)}-\frac{r}{r(r+1)}ight]$ $=2 \sum_{r=1}^{10}\left(\frac{1}{r}-\frac{1}{r+1}ight)$ $=2\left[\left(\frac{1}{1}-\frac{1}{2}ight)+\left(\frac{1}{2}-\frac{1}{3}ight)+\left(\frac{1}{3}-\frac{1}{4}ight)+\ldots+\left(\frac{1}{10}-\frac{1}{11}ight)ight]$ $=2\left[1-\frac{1}{11}ight]=2 	imes \frac{10}{11}=\frac{20}{11}$

The sum of the series: $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots \ldots$. upto 10 terms, is:

Solution